Question

Prove tan^-1\((\frac{√1+x-√1-x}{√1+x+√1-x}=\frac{π}{4}-\frac{1}{2}cos^-1x,-\frac{1}{√2}≤x≤1.\)[Hint: putx = cos 2θ]

Answer 1

Put x=cos2θ so that θ=\(\frac{1}{2}cos-1\)x.Then we have:
\(LHS=tan^-1(\frac{√1+x-√1-x}{√1+x+√1-x)}\)
\(=tan^-1(\frac{√1+cos2θ-√1-cos2θ}{√1+cos2θ+√1-cos2θ})\)
\(=tan^-1(\frac{√2cos^2θ-√2sin^2θ}{√2cos^2θ+√2sin^2θ)}\)
\(=tan^-1(\frac{√2cosθ-√2sinθ}{√2cosθ+√2sinθ})\)
\(=tan^-1(\frac{cosθ-sinθ}{cosθ+sinθ})=tan-1(\frac{1-tanθ}{1+tanθ})\)
\(=tan^-1(1)-tan^-1(tanθ)= \frac{π}{4}-θ=\frac{π}{4}-\frac{1}{2}cos^-1x=RHS\)